integral of 1/(1+(1+x^2)^{1/2)}

解答

\int \frac{1}{1 + ( 1 + x ^{2} ) ^{\frac{1}{2}}} d x

ln tan (\frac{arctan ( x )}{2}) + 1 - ln tan (\frac{arctan ( x )}{2}) - 1 - tan (\frac{arctan ( x )}{2}) + C

求解步骤

\int \frac{1}{1 + ( 1 + x ^{2} ) ^{\frac{1}{2}}} d x

使用三角换元法

= \int \frac{sec ^{2} ( u )}{1 + sec ( u )} d u

化简

\frac{sec ^{2} ( u )}{1 + sec ( u )} : \frac{1}{cos ( u ) + cos ^{2} ( u )}

= \int \frac{1}{cos ( u ) + cos ^{2} ( u )} d u

使用换元积分法

= \int \frac{v ^{2} + 1}{- v ^{2} + 1} d v

\frac{v ^{2} + 1}{- v ^{2} + 1} = \frac{2}{- v ^{2} + 1} - 1

= \int \frac{2}{- v ^{2} + 1} - 1 d v

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2}{- v ^{2} + 1} d v - \int 1 d v

\int \frac{2}{- v ^{2} + 1} d v = ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣

\int 1 d v = v

= ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣ - v

代回

= ln tan (\frac{arctan ( x )}{2}) + 1 - ln tan (\frac{arctan ( x )}{2}) - 1 - tan (\frac{arctan ( x )}{2})

解答补常数

= ln tan (\frac{arctan ( x )}{2}) + 1 - ln tan (\frac{arctan ( x )}{2}) - 1 - tan (\frac{arctan ( x )}{2}) + C

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